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anonymous
 4 years ago
Important Question. Please explain with steps.
anonymous
 4 years ago
Important Question. Please explain with steps.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Mani_Jha , @saifoo.khan , @heena , @supercrazy92 please help.

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.0@Mani_Jha is a boss at chem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, that's why i called Mani Bhai also. He is my teacher. I hope he will do it this time too.:)))))

Mani_Jha
 4 years ago
Best ResponseYou've already chosen the best response.1Let the concentration of NH3 be x. Now, the resultant solution was basic(as it had to be neutralized by NaOH). During the reaction of neutralization: M1V1=M2V2 x20=30*0.1 Find z. Let the initial volume of NH3 be V. Then when NH3 reacted initially with H2S04, 100 ml of H2S04 neutralized (v20) ml of NH3(Because 20 ml was unneutralized and made the solution basic) Again use the law of equivalents in this case(be careful: the nfactor of H2S04 is 2) Did you understand? @Saifoo.khan, you just guessed that, didnt u?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Mani Bhai, can u show the second equation what we need to take?

Mani_Jha
 4 years ago
Best ResponseYou've already chosen the best response.1(v20)x=100*0.1*2. ok?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, mani bhaiiiiiii

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wordings of these problems are terrible; teachers should bear in mind that students are not reading their minds and that wordings should be unambiguous. In first problem, I find 17 millimole NH3 or 0.38 litre in STP. That holds if "equivalence" in the wording is meant for 'first equivalence', because eventually, NH4+ formed will also be titrated. In second problem there should be 1.5 millimole of carbonate and 0.5 millimole of hydrogencarbonate. Here again, problem should read: "The second end point was reached when AN EXTRA 20 ml of the acid was added

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oooookkkkkk, I will consult with my teacher about this. But these r the exact wordings that our teacher gave.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What about the answers?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0btw, do you use 22.4 litres/mole or 22.7 litres/mole for molar volume in STP?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you have expected answers ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not now. i need to submit it as my assignment. i couldn't do it, so i asked it here.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do my answers coincide with yours of ManiJha's? Do you understand what reactions are at stake here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My teacher gave the answer just now as 112 ml for the first one, and for the second one: Weight of NaOH = 0.8g, weight of Na2CO3 = 1.59 g

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01st one is correct. I forgot to divide something by 5, as you take only a 20 mL sample of the 100 mL solution. So there were 5 millimoles of ammonia ie a volume of 112 mL

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have 159 mg of Na2CO3 which is not 1.59 g, unless concentration af acid is 1M instead of 0.1M. Then I do not understand the NaOH part. Your solution is supposed to have NaHCO3. What do you think?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Even I think the same.
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